After looking at additive and multiplicative persistence, I wanted to look at a possible exponent extension.
I decided the most logical procession that left a "root" would be to take each digit to the power of it's position such that
$E_1(982)=9^3+8^2+2^1=795$
$E_2(982)=E_1(795)=7^3+9^2+5^1=429$ $\cdots E_7(982)=1$
Repeatedly until one digit remains gives the root, similar to the digital root.
$E_n(982)=1$
And the number of steps it takes to reach a one digit answer, or the persistence.
$E_p(982)=7$
Iterating this until a single digit remains takes 7 steps and results in the number 1.
Below is a table of the smallest integer k, with persistence n
$$ \left[ \begin{array}{cc} n&k\\ 0&1\\ 1&10 \\ 2&19\\ 3&43 \\ 4&67 \\ 5&83 \\ 6&92 \\ 7&293 \\ 8&691 \\ 9&958 \\ 10&7849 \\ 11&49670 \\ 12&94976 \\ 13&880096 \\ 14&7090761 \\ \end{array} \right] $$
Also $E_n(k)=1$ at least for the table above, but is this true for all k?
Unfortunately my limited python skills leads to a pretty inefficient search.
So my questions would be.
Is there a function that solves for k? What is the probability of getting a certain digit as the root? What about persistence?